Javascript remove all occurrence of duplicate element, leaving the only one that is unique

Question

I want to remove elements that occurr more than once and get the unique element. The array always has 3 elements. Lets say i have an array [2,3,2], then I need to get 3 which is only unique in the array(removing both 2s, because they occur more than once).

I have tried with following code, but surely it doesnot work as expected.

var firstArrTemp = [2,3,2];
var sorted_arr = firstArrTemp.sort();
var unique_element;
for (var i = 0; i < sorted_arr.length - 1; i++) {
    if (sorted_arr[i + 1] != sorted_arr[i]) {
        unique_element=sorted_arr[i];
    }
}

alert(unique_element);

Thanks!

Answer 1

This should do the trick:

Array.prototype.getUnique = function(){
    var uniques = [];
    for(var i = 0, l = this.length; i < l; ++i){
        if(this.lastIndexOf(this[i]) == this.indexOf(this[i])) {
            uniques.push(this[i]);
        }
    }
    return uniques;
}

// Usage:

var a = [2, 6, 7856, 24, 6, 24];
alert(JSON.stringify(a.getUnique()));

console.log(a.getUnique()); // [2, 7856]

To check if a specific item is unique in the array, it just checks if the first index it's found at, matches the last index it's found at.

Answer 2

One alternative with filter() function:

var myArray = [1,2,3,2,2,4,3,7,3].sort();
var uniqueValues = myArray.filter(function(item, i, arr) {
  return (item !== arr[i-1] && item !== arr[i+1]);
});

Where uniqueValues = [1,4,7]